trig functions and the chain rule - texas a&m...
TRANSCRIPT
The important differentiation formulas fortrigonometric functions are
d
dxsinx = cosx,
d
dxcosx = − sinx.
Memorize them! To evaluate any other trig deriva-tive, you just combine these with the product (andquotient) rule and chain rule and the definitions ofthe other trig functions, of which the most impor-tant is
tanx =sin x
cos x.
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Exercise 3.4.19
Prove that
d
dxcotx = − csc2 x.
Exercise 3.4.23
Find the derivative of y = cscx cotx.
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What is ddx cotx ? Well, the definition of the
cotangent is
cotx =cosx
sinx.
So, by the quotient rule, its derivative is
sinx(− sinx)− cosx(cosx)
sin2 x
= −1
sin2 x≡ − csc2 x
(since sin2 x+ cos2 x = 1).
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To differentiate cscx cotx use the productrule:
dy
dx=
d cscx
dxcotx+ cscx
d cotx
dx.
The second derivative is the one we just cal-culated, and the other one is found similarly(Ex. 3.4.17):
d
dxcscx = − cscx cotx.
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Sody
dx= − cscx cot2 x− csc3 x.
This could be rewritten using trig identities, butthe other versions are no simpler.
Another method:
y = csc x cot x =cos x
sin2 x.
Now use the quotient rule (and cancel some extrafactors of sin x as the last step).
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We want to find the functions whose deriva-tive is h(x) = sinx − 2 cosx. If we know twofunctions whose derivatives are (respectively)sinx and cosx, we’re home free. But we do!
d(− cosx)
dx= sinx,
d sinx
dx= cosx.
So we let H(x) = − cosx− 2 sinx and check thatH ′(x) = h(x). The most general antiderivative ofh is H(x) + C where C is an arbitrary constant.
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Now let’s drop back to see where the trigderivatives came from. On pp. 180–181 we’reoffered 4 trigonometric limits, but they are not ofequal profundity.
The first two just say that the sine and co-sine functions are continuous at θ = 0. (But youalready knew that, didn’t you?)
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The third limit is the important one:
limθ→0
sin θ
θ= 1.
It can be proved from the inequalities
sin θ < θ < tan θ (for 0 < θ < π/2),
which are made obvious by drawing some pic-tures.
It says that sin θ “behaves like” θ when θ issmall.
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In contrast, the fourth limit formula,
limθ→0
cos θ − 1
θ= 0,
says that cos θ “behaves like” 1 and that the dif-ference from 1 vanishes faster than θ as θ goesto 0.
In fact, later we will see that
cos θ ≈ 1−θ2
2.
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Split the function into a product of functionswhose limits we know:
tan 3x
3 tan 2x=
1
3
sin 3x
cos 3x
cos 2x
sin 2x
=sin 3x
3x
2x
sin 2x
cos 2x
2 cos 3x.
As x → 0, 2x and 3x approach 0 as well. There-fore, the two sine quotients approach 1. Eachcosine also goes to 1. So the limit is 1
2 .
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The chain rule is the most important andpowerful theorem about derivatives. For a firstlook at it, let’s approach the last example of lastweek’s lecture in a different way:
Exercise 3.3.11 (revisited and shortened)
A stone is dropped into a lake, creating a circu-lar ripple that travels outward at a speed of 60cm/s. Find the rate at which the circle’s area isincreasing after 3 s .
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In applied problems it’s usually easier to usethe “Leibniz notation”, such as df/dx, instead ofthe “prime” notation for derivatives (which is es-sentially Newton’s notation).
The area of a circle of radius r is
A = πr2.
SodA
dr= 2πr.
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(Notice that dA/dr = 2πr is the circumference.That makes sense, since when the radius changesby ∆r, the region enclosed changes by a thin cir-cular strip of length 2πr and width ∆r, hence area∆A = 2πr∆r.)
From A′(r) = 2πr we could compute therate of change of area with respect to radius
by plugging in the appropriate value of r(namely, 60× 3 = 180). But the question asks forthe rate with respect to time and tells us that
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dr
dt= 60 cm/s .
Common sense says that we should just multiplyA′(r) by 60, getting
dA
dt=
dA
dr
dr
dt= 2π(60)2t, t = 3.
Of course, this is the same result we got lastweek.
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Now consider a slight variation on the prob-lem:
Exercise 3.3.11 (modified)
A stone is dropped into a lake, creating a circu-lar ripple that travels outward at a speed of 60cm/s. Find the rate at which the area is increas-ing when the radius is 180 cm .
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To answer this question by the method oflast week, using A(t) = 3600πt2, we would needto calculate the time when r = 180. That’s easyenough in this case (t = 3), but it carries the ar-gument through an unnecessary loop through aninverse function. It is more natural and simplerto use this week’s formula,
dA
dt=
dA
dr
dr
dt.
It givesdA
dt= (2πr)60 = 21600π.
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In both versions of the exercise we dealtwith a function of the type A(r(t)), where theoutput of one function is plugged in as the in-put to a different one. The composite function
is sometimes denoted A ◦ r (or (A ◦ r)(t)). Thechain rule says that
(A ◦ r)′(t) = A′(r(t))r′(t),
ordA
dt=
dA
dr
dr
dt.
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Now we can use the chain rule to differenti-ate particular functions.
Exercise 3.5.7
Differentiate
G(x) = (3x− 2)10(5x2 − x+ 1)12.
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G(x) = (3x− 2)10(5x2 − x+ 1)12.
It would be foolish to multiply out the pow-ers when we can use the chain rule instead. Ofcourse, the first step is a product rule.
G′(x) = 10(3x − 2)
9 d
dx(3x − 2) (5x
2− x + 1)
12
+12(3x − 2)10
(5x2− x + 1)
11 d
dx(5x
2− x + 1)
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= 30(3x − 2)9(5x2− x + 1)12
+ 12(10x − 1)(3x − 2)10
(5x2− x + 1)
11.
(The book’s answer combines some terms at theexpense of factoring out a messy polynomial.)
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Note that it is not smart to use the quotientrule on a problem like
d
dx
x+ 1
(x2 + 1)3.
You’ll find yourself cancelling extra factors ofx2 + 1. It’s much better to use the product ruleon
(x+ 1)(x2 + 1)−3,
getting only 4 factors of (x2 + 1)−1 instead of 6.
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y = 8(4 + 3x)−1/2.
dy
dx= −4(4 + 3x)−3/2(3).
When x = 4, y′ = −12(16)−3/2 = −3/16 (andy = 8(16)−1/2 = 2 as the problem claims). So thetangent is
y = 2−3
16(x− 4).
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The book simplifies the equation
y = 2 −
3
16(x − 4)
to 3x + 16y = 44, but I think it is better to leavesuch equations in the form that emphasizes thedependence on ∆x (= x − 4 in this case). Thepoint of a tangent line is that is the best linearapproximation to the curve in the region where ∆xis small.
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Exercise 3.5.59
Suppose F (x) = f(g(x)) and
g(3) = 6, g′(3) = 4, f ′(3) = 2, f ′(6) = 7.
Find F ′(3).
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